Induction: $n + 3 < n!$ for all $n>3$

Question

I have a proof that I am trying to prove and I am getting stuck at the inductive hypothesis. This is my theorem:

For all real numbers $n>3$, the following is true: $n + 3 < n!$.

I have proven true for $n = 4$, and will assume true for some arbitrary value $k$, i.e.,

$$k + 3 < n!,$$

and I want to prove for $k+1$, i.e.,

$$(k+1) + 3 < (k+1)!.$$

Consider the $k+1$ term:

$$(k+1)+3 = ?$$

I am confused on how to approach the next step.

Ok here is how I am proceeding. It seems really long so if anyone has a better way let me know: $$ =(k+3)+1 $$ $$ <(k!)+1 $$ $$ <k!+k! $$ $$ =2k! $$ $$ <(k+1)k! $$ $$ =(k+1)! $$ Therefore both sides are equivalent.

Answer 1

As you are trying to solve this problem, I'll only give you a hint.

Inductive Step: we want to show $(n+1)+3 < (n+1)!$

That's equivalent to $n+4 < (n+1)\cdot n!$ by the property of the factorial.

We can distribute: $n+4 < (n\cdot n!) + (1\cdot n!)$

Can you take it from here?

Answer 2

Since by induction hypothesis,

$$k+3< k!$$

for $k>4$, multiply both sides by $(k+1)$ to get

$$(k+3)(k+1) < k! (k+1)$$

or

$$(k+3)(k+1) < (k+1)!$$

I'll leave the rest for you to think about, as a hint, remember that's an inequality.

Answer 3

Suppose $k! \gt k+3$ is true:

\begin{align*} \ (k+1)! &=k!\cdot(k+1) \\ &\gt(k+3)(k+1) \\ &=k^2+4k+3 \\ &\gt k^2+k+3 \\ &\gt (k+1)+3\ldots(\text{where}\space k\gt 3) \end{align*}

Answer 4

Induction is overkill here.

For $n \gt 3$ we have that $n+3 \lt n + n = 2\cdot n \lt 1\cdot 2 \cdot3\cdots (n-1) \cdot n =n!$

Answer 5

Suppose $n+3< n!$ and $(n+1)!\leq n+1+3$ , we see the following:

$$(n+1)!\leq n+1+3 \Rightarrow (n+1)n!\leq n+1+3 $$

$$\Rightarrow (n+1)(n+3)\leq(n+1)n!\leq n+1+3 $$

$$(n+1)(n+3)\leq (n+1)+3 \Rightarrow (n+1)(n+3)-(n+1) \leq 3$$

$$ \Rightarrow (n+1)(n+3-1)\leq 3$$

$$(n+1)(n+2)\leq 3$$

This should strike some thing s wrong... what is that?